Single-Variable-Calculus · Unit 14 · Video 2 · Interactive Practice

Area Under y = x²: Squeezing the Sum of Squares

IKey Formulas

Formula	Name	Where it comes from
$\tfrac{1}{3}n^3 < 1^2 + 2^2 + \cdots + n^2 < \tfrac{1}{3}(n+1)^3$	Pyramid bounds	Inner pyramid (base $n$ , height $n$ ) and outer pyramid (base $n+1$ , height $n+1$ )
$\tfrac{1}{3} < \dfrac{1^2 + 2^2 + \cdots + n^2}{n^3} < \tfrac{1}{3}\left(1 + \tfrac{1}{n}\right)^3$	Squeeze on the ratio	Divide all three parts by $n^3$
$\displaystyle\lim_{n\to\infty} \frac{1^2 + 2^2 + \cdots + n^2}{n^3} = \frac{1}{3}$	Limit of the ratio	Both bounds tend to $\tfrac{1}{3}$
$\displaystyle\int_0^b x^2\,dx = \lim_{n\to\infty} \frac{b^3}{n^3}\left(1^2 + \cdots + n^2\right) = \frac{b^3}{3}$	Area under the parabola	Rectangle sums with $b$ fixed

Key Insight: The pyramids never give the exact sum of squares; they only show it grows like $\tfrac{1}{3}n^3$ , and that is all the area needs.

IIVisualization 1 — The Staircase Between Two Pyramids

A staircase of $n$ levels holds $1^2 + 2^2 + \cdots + n^2$ blocks and sits between two solid pyramids.

Levels n 4 Show

💡 The picture is a side view; in three dimensions the levels are square slabs of $n^2, (n-1)^2, \ldots, 1^2$ cubes, and each solid pyramid has volume $\tfrac{1}{3}(\text{base area})(\text{height})$ .

IIIVisualization 2 — The Squeeze on the Ratio

After dividing by $n^3$ , the ratio is trapped between $\tfrac{1}{3}$ and a bound that shrinks to $\tfrac{1}{3}$ .

n 4

IVVisualization 3 — Back to the Rectangles

With $b$ fixed, the rectangle sum is $b^3$ times the ratio, so it tends to $\tfrac{b^3}{3}$ .

Rectangles n 8 Endpoint b 1

VQuiz Questions

Problem 1 · The Bounds for Six Levels

Given: A staircase pyramid with $n = 6$ levels holds $1^2 + 2^2 + \cdots + 6^2 = 91$ blocks. Which inequality do the inner and outer pyramids give?

✅ Correct!

\tfrac{1}{3}\cdot 6^3 = 72

and

\tfrac{1}{3}\cdot 7^3 = \tfrac{343}{3} \approx 114.3

, and

91

lies between them.

❌ Missing the $\tfrac{1}{3}$ . A pyramid's volume is

\tfrac{1}{3}(\text{base area})(\text{height})

, not base area times height;

216 < 91

is false anyway.

❌ Shifted by one. The inner pyramid has base

n = 6

and the outer one base

n + 1 = 7

; you used

5

and

6

, and then

91 < 72

fails.

❌ Those are areas, not volumes. The bounds are cubes:

\tfrac{1}{3}n^2 \cdot n

, since the base area is

n^2

and the height is

n

Show solution

The inner pyramid has an $n \times n$ base and height $n$ , so its volume is $\tfrac{1}{3}n^2\cdot n = \tfrac{1}{3}n^3$ . The outer pyramid has an $(n+1)\times(n+1)$ base and height $n+1$ , so its volume is $\tfrac{1}{3}(n+1)^3$ .

\tfrac{1}{3}\cdot 6^3 = \tfrac{216}{3} = 72, \qquad \tfrac{1}{3}\cdot 7^3 = \tfrac{343}{3} \approx 114.3

The staircase sits between them, so $72 < 91 < \tfrac{343}{3}$ .

Problem 2 · What the Squeeze Proves

Given: $\tfrac{1}{3}n^3 < 1^2 + 2^2 + \cdots + n^2 < \tfrac{1}{3}(n+1)^3$ for every $n$ . Write $S_n = 1^2 + 2^2 + \cdots + n^2$ . Which statement is true?

✅ Correct! The ratio is squeezed to

\tfrac{1}{3}

, while the gap grows: about

8.7

n = 4

51.7

n = 10

5017

n = 100

❌ The gap does not shrink. At

n = 10

S_{10} - \tfrac{1000}{3} = 385 - 333.3 \approx 51.7

; at

n = 100

it is about

5017

. Only the ratio settles down.

❌ Both bounds are needed. The lower bound only keeps the ratio above

\tfrac{1}{3}

; it is the upper bound

\tfrac{1}{3}(1 + \tfrac{1}{n})^3 \to \tfrac{1}{3}

that forces it down.

❌ The lower bound rules that out. Dividing

\tfrac{1}{3}n^3 < S_n

n^3

gives

\dfrac{S_n}{n^3} > \tfrac{1}{3}

for every

n

, so the ratio cannot tend to

0

Show solution

Divide all three parts by $n^3$ :

\frac{1}{3} < \frac{S_n}{n^3} < \frac{1}{3}\left(1 + \frac{1}{n}\right)^3.

As $n \to \infty$ , $\tfrac{1}{n} \to 0$ , so the upper bound tends to $\tfrac{1}{3}$ ; the lower bound is already $\tfrac{1}{3}$ . The ratio is squeezed: $\dfrac{S_n}{n^3} \to \tfrac{1}{3}$ .

The difference is another matter: $S_4 - \tfrac{64}{3} = 30 - 21.33 \approx 8.7$ , $S_{10} - \tfrac{1000}{3} \approx 51.7$ , $S_{100} - \tfrac{10^6}{3} \approx 5017$ . The gap grows, but more slowly than $n^3$ , so it vanishes only after dividing by $n^3$ .

Problem 3 · A Rectangle Sum and Its Limit

Given: The region under $y = x^2$ from $0$ to $b = 2$ , cut into $n = 4$ strips of equal width, each rectangle's height taken at its right endpoint. Find the rectangle sum, then the exact area.

Rectangle sum for n = 4

Exact area under y = x² from 0 to 2

✅ Correct!

\tfrac{8}{64}\cdot 30 = \tfrac{15}{4}

overestimates; the limit is

\tfrac{2^3}{3} = \tfrac{8}{3}

❌ Check the rectangle sum. The widths are

\tfrac{b}{n} = \tfrac{1}{2}

and the heights are

\left(\tfrac{kb}{n}\right)^2

for

k = 1, \ldots, 4

, which gives the factor

\tfrac{b^3}{n^3}

in front of

1^2 + 2^2 + 3^2 + 4^2

❌ Check the limit. The area is the limit of

b^3\cdot\tfrac{S_n}{n^3}

, and

\tfrac{S_n}{n^3} \to \tfrac{1}{3}

, so it is

\tfrac{1}{3}

b^3 = 8

: keep the

\tfrac{1}{3}

, use

b^3

rather than

b^2

, and do not stop at one rectangle sum.

Show solution

Each strip has width $\tfrac{2}{4} = \tfrac{1}{2}$ ; the right endpoints are $\tfrac{1}{2}, 1, \tfrac{3}{2}, 2$ .

\tfrac{1}{2}\left(\tfrac{1}{4} + 1 + \tfrac{9}{4} + 4\right) = \tfrac{1}{2}\cdot\tfrac{30}{4} = \tfrac{15}{4}

Equivalently, $\dfrac{b^3}{n^3}\left(1^2 + 2^2 + 3^2 + 4^2\right) = \dfrac{8}{64}\cdot 30 = \dfrac{15}{4}$ .

As $n \to \infty$ with $b = 2$ fixed, $\dfrac{b^3}{n^3}S_n = b^3\cdot\dfrac{S_n}{n^3} \to 8\cdot\tfrac{1}{3} = \tfrac{8}{3}$ .

The left-endpoint sum $\tfrac{8}{64}(0 + 1 + 4 + 9) = \tfrac{7}{4}$ underestimates; $\tfrac{8}{3} \approx 2.67$ lies between $\tfrac{7}{4}$ and $\tfrac{15}{4}$ .

Problem 4 · Area That Does Not Start at Zero

Given: $\displaystyle\int_0^b x^2\,dx = \frac{b^3}{3}$ . Find the area under $y = x^2$ between $x = 1$ and $x = 2$ .

✅ Correct! Area from

0

2

minus area from

0

1

\tfrac{8}{3} - \tfrac{1}{3} = \tfrac{7}{3}

❌ That is the area from $0$ to $2$ . It includes the piece from

0

1

, which has area

\tfrac{1}{3}

and must be removed.

❌ The formula is not $\tfrac{(b-a)^3}{3}$ . The interval

[1, 2]

has length

1

, but the parabola is much higher there than on

[0, 1]

❌ That is $2^2 - 1^2$ . Areas under

y = x^2

use cubes: subtract

\tfrac{1^3}{3}

from

\tfrac{2^3}{3}

Show solution

The region from $0$ to $2$ splits into the piece from $0$ to $1$ and the piece from $1$ to $2$ , so

\text{area from 1 to 2} = \frac{2^3}{3} - \frac{1^3}{3} = \frac{8}{3} - \frac{1}{3} = \frac{7}{3}.

Sanity check: on $[1, 2]$ the height runs from $1$ to $4$ over width $1$ , so the area must lie between $1$ and $4$ ; $\tfrac{7}{3} \approx 2.33$ does.

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